Question

derive ME law of conservation of momentum ​

Answer 1

$\text{\large{\underline{\underline{Conservation of Momentum}}}}$

The law of conservation of momentum means that whenever one body gives Momentum then some other body must laws and equal amount of momentum .

$\text{\large{\underline{\underline{Derivation}}}}$

$\textsf{ \underline{First :}}$

When the force $F_1$ of a body acts on the another body for a time $t$ , then the velocity of another body changes from $u_2$ to $v_2$ . So,

$\text{Acceleration of another body, } \\ \text{a}_2 = \frac{( \text{v}_2 -\text{u}_2 ) }{\text{t}}$

But, force = mass × acceleration, So the force $F_1$ exerted by the body on the another body is given by :

$\text{F}_1 = \text{m}_2 \times \frac{( \text{v}_2 - \text{u}_2)}{ \text{t}} - - - - - (1)$

$\textsf{ \underline{Second :}}$

When the force $F_1$ of another body acts on the body for a time $t$ , then the velocity of the body changes from $v_2$ to $u_2$ . So,

$\text{Acceleration of another body, } \\ \text{a}_1 = \frac{( \text{v}_1 -\text{u}_1 ) }{\text{t}}$

But, force = mass × acceleration, So the force $F_2$ exerted by the body on the body is given by :

$\text{F}_2 = \text{m}_1 \times \frac{( \text{v}_1 - \text{u}_1)}{ \text{t}}$

Now, the force $F_2$ exerted by the body is the $\textsf {action}$ of the force $F_1$ exerted by the another body is the $\textsf {reaction}$. But according to the third law of motion, the action and reaction are equal and opposite . So,

$\text{F}_1 = - \text{F}_2$

$\text{now \: putting \: the \: value \: of \: } \\ \text{F}_1 \text{and } \text{F}_2 \: \text{from \: eq. (1) and (2)}$

$\text{m}_2 \times \frac{( \text{v}_2 - \text{u}_2)}{ \text{t}} = - \text{m}_1 \times \frac{( \text{v}_1 - \text{u}_1)}{ \text{t}} \\ \\ \text{m}_2 \times \frac{( \text{v}_2 - \text{u}_2)}{ \cancel { \text{t}}} = - \text{m}_1 \times \frac{( \text{v}_1 - \text{u}_1)}{ \cancel{ \text{t}}}$

Cancelling t from both sides, we get :

$\: \: \: \: \: \text{m}_2( \text{v}_2 - \text{u}_2) \\ \\ \implies \: \text{m}_2\text{v}_2 - \text{m}_2\text{u}_2 = - \text{m}_1\text{v}_1 + \text{m}_1\text{u}_1 \\ \\ \implies \: \text{m}_2\text{v}_2 + \text{m}_1\text{v}_1 = \text{m}_2\text{u}_2 + \text{m}_1\text{u}_1 \\ \\ \implies \: \text{m}_1\text{u}_1 + \text{m}_2\text{u}_2 = \text{m}_1\text{v}_1 + \text{m}_2\text{v}_2$

Answer 2

Explanation:

Law of conservation of momentum states that when two objects collide with each other , the sum of their linear momentum always remains same or we can say conserved and is not effected by any action, reaction only in case is no external unbalanced force is applied on the bodies.

DERIVING THE LAW OF CONSERVATION OF MOMENTUM-

Let,

mA = Mass of ball A

mB= Mass of ball Ba

uA= initial velocity of ball A

uB= initial velocity of ball B

vA= Velocity after collision of ball A

vB= Velocity after collision of ball B

Fab= Force exerted by A on B

Fba= Force exerted by B on A

Now,

Change in momentum of A= momentum of A after collision - momentum of A before collision

= mA vA - mA uA

Rate of change of momentum A= Change in momentum of A/ time taken

= mA vA - mA uA/t

Force exerted by B on A (Fba)=

Fba= mA vA - mA uA/t. [i]

In the same way,

Rate of change of momentum of B=

mV vB - mB uB/t

Force exerted by A on B (Fab)=

Fab= mB vB - mB uB/t. [ii]

Newton's third law of motion states that every action has an equal and opposite reaction, then,

Fab= -Fba [ ' -- ' sign is used to indicate that 1 object is moving in opposite direction after collision]

Using [i] and [ii] , we have

mB vB - mB uB/t = - (mA vA - mA uA/t)

mB vB - mB uB= - mA vA + mA uA

Finally we get,

mB vB + mA vA = mB uB + mA uA

This is the derivation of conservation of linear momentum.

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